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Binomial expansion II

Binomial expansion: (a+bx)^n

Binomial expansion: (a+bx)^n

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Summary

Binomial expansion: (a+bx)n(a+bx)^n(a+bx)n​​

In a nutshell

You can use the formula for the expansion of (1+x)n(1+x)^n(1+x)n​ to compute the expansion of (a+bx)n(a+bx)^n(a+bx)n for any constants aaa​ and bbb​. In the case that ​nnn is not a non-negative integer, this expansion will only be valid when ∣x∣<∣ab∣|x| < \left| \dfrac{a}{b}\right|∣x∣<​ba​​​.



Using the expansion of (1+x)n(1+x)^n(1+x)n​​

The expansion formula tells you that:


​(1+x)n=1+nx+n(n−1)2!x2+⋯+(nr)xr+…(1+x)^n = 1 + nx + \dfrac{n(n-1)}{2!}x^2 + \dots + \binom{n}{r}x^r + \dots(1+x)n=1+nx+2!n(n−1)​x2+⋯+(rn​)xr+…​​


This holds in general whenever nnn​ is a non-negative integer, as in these cases only finitely many binomial coefficients are nonzero, whereas when nnn​ is not a non-negative integer this produces an infinite series, and only holds whenever ∣x∣<1|x| < 1∣x∣<1​.


You can use this to derive the binomial expansion for (a+bx)n(a+bx)^n(a+bx)n​ using this formula. If a=0a=0a=0​, then (a+bx)n=bnxn(a+bx)^n = b^nx^n(a+bx)n=bnxn​, but otherwise:


​(a+bx)n=(a(1+bax))n=an(1+bax)n=an[1+n(bax)+n(n−1)2!(bax)2+⋯+n(n−1)(n−2)...(n−r+1)r!(bax)n+… ]\begin{aligned}(a + bx)^n &= \left(a\left(1+\dfrac{b}{a}x\right)\right)^n\\&= a^n\left(1 + \dfrac{b}{a}x\right)^n \\&= a^n\left[1 + n\left(\dfrac{b}{a}x\right) + \dfrac{n(n-1)}{2!}\left(\dfrac{b}{a}x\right)^2 + \dots + \dfrac{n(n-1)(n-2)...(n-r+1)}{r!}\left(\dfrac{b}{a}x\right)^n + \dots\right]\end{aligned}(a+bx)n​=(a(1+ab​x))n=an(1+ab​x)n=an[1+n(ab​x)+2!n(n−1)​(ab​x)2+⋯+r!n(n−1)(n−2)...(n−r+1)​(ab​x)n+…]​​​


This holds in general whenever n∈Nn \in \mathbb{N}n∈N​, and if nnn​ is any other real value this only holds when ∣bax∣<1\left|\dfrac{b}{a}x\right| < 1​ab​x​<1​, i.e. when ∣x∣<∣ab∣|x| < \left|\dfrac{a}{b}\right|∣x∣<​ba​​​.



Example 1

Find the first 444​ terms in the expansion of 112−3x\dfrac{1}{\sqrt{12 - 3x}}12−3x​1​​, and determine the range of values of xxx for which this expansion holds.


Rearrange 112−3x\dfrac{1}{\sqrt{12-3x}}12−3x​1​:


112−3x=(12−3x)−1=(12−3x)−12=(12)−12(1−14x)−12\begin{aligned}\dfrac{1}{\sqrt{12 - 3x}} &= (\sqrt{12-3x})^{-1}\\&= \left(12-3x\right)^{-\dfrac12}\\&= (12)^{-\dfrac12}\left(1 - \dfrac14x\right)^{-\dfrac12}\end{aligned}12−3x​1​​=(12−3x​)−1=(12−3x)−21​=(12)−21​(1−41​x)−21​​​​

​

Compute the first 444 terms using the expansion of (1−14x)−12\left(1 - \dfrac14x\right)^{-\dfrac12}(1−41​x)−21​:


​112−3x=(12)−12(1−14x)−12=(12)−12[1+(−12)(−14x)+(−12)(−32)2!(−14x)2+(−12)(−32)(−52)3!(−14x)3+… ]=(12)−12[1+(−1)22×4x+(−1)4×323×42x2+(−1)6×3×523×6×43x3+… ]=36[1+18x+3128x2+153072x3+… ]=36+348x+3256x2+536144x3+…\begin{aligned}\dfrac{1}{\sqrt{12-3x}} &= (12)^{-\dfrac12}\left(1-\dfrac14x\right)^{-\dfrac12}\\&=(12)^{-\dfrac12}\left[1 +\left(-\dfrac12\right)\left(- \dfrac14x\right) + \dfrac{\left(-\dfrac12\right) \left(-\dfrac32\right)}{2!}\left(-\dfrac14x\right)^2 + \dfrac{\left(-\dfrac12\right)\left(-\dfrac32\right)\left(-\dfrac52\right)}{3!}\left(-\dfrac14x\right)^3 + \dots\right]\\&= (12)^{-\dfrac12}\left[1 + \dfrac{(-1)^2}{2\times 4}x + \dfrac{(-1)^4\times3}{2^3 \times 4^2}x^2 + \dfrac{(-1)^6 \times 3 \times 5}{2^3 \times 6 \times 4^3}x^3 + \dots\right]\\&=\dfrac{\sqrt{3}}{6}\left[1 + \dfrac{1}{8}x + \dfrac{3}{128}x^2 + \dfrac{15}{3072}x^3 + \dots\right]\\&= \dfrac{\sqrt{3}}{6} + \dfrac{\sqrt{3}}{48}x + \dfrac{\sqrt{3}}{256}x^2 + \dfrac{5\sqrt{3}}{6144}x^3 + \dots\end{aligned}12−3x​1​​=(12)−21​(1−41​x)−21​=(12)−21​​1+(−21​)(−41​x)+2!(−21​)(−23​)​(−41​x)2+3!(−21​)(−23​)(−25​)​(−41​x)3+…​=(12)−21​[1+2×4(−1)2​x+23×42(−1)4×3​x2+23×6×43(−1)6×3×5​x3+…]=63​​[1+81​x+1283​x2+307215​x3+…]=63​​+483​​x+2563​​x2+614453​​x3+…​​​


This expansion is valid when ∣−14x∣<1\left|-\dfrac14x\right| < 1​−41​x​<1, i.e. when ∣x∣<4|x| < 4∣x∣<4.


Therefore, the first four terms of the expansion of 112−3x\dfrac{1}{\sqrt{12-3x}}12−3x​1​ are 36+348x+3256x2+536144x3+…‾\underline{\dfrac{\sqrt{3}}{6} + \dfrac{\sqrt{3}}{48}x + \dfrac{\sqrt{3}}{256}x^2 + \dfrac{5\sqrt{3}}{6144}x^3 + \dots}63​​+483​​x+2563​​x2+614453​​x3+…​, and the range of values of xxx for which this expansion holds is ∣x∣<4‾\underline{|x|< 4}∣x∣<4​.


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FAQs - Frequently Asked Questions

For what values of n is the expansion of (a+bx)^n an infinite series?

When n is not a non-negative integer.

For what range of values is the expansion of (a+bx)^n valid?

In the case that n is not a natural number, this expansion will only be valid when |x| < |a/b|.

How do I expand (a+bx)^n?

You can use the formula for the expansion of (1+x)^n to compute the expansion for (a+bx)^n for any constants a and b.

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